Generate a random number from 1 to 100.
If it is less than or equal to x, return true, else return false.
This will ensure that ur returning true with x/100 probability.
Cheers
Nikhil Jindal
On Thu, Jul 28, 2011 at 4:21 PM, KK kunalkapadi...@gmail.com wrote:
bool foo(int x)
// Implement
Anshu here has given a Perfect soln!
Sunny's code is also correct but is a bit less efficient than anshu's.
Cheers
Nikhil Jindal
http://sites.google.com/site/aboutnikhiljindal/
On Fri, May 27, 2011 at 9:11 PM, anshu mishra anshumishra6...@gmail.comwrote:
@all go through this code
, not just the first digit but the complete number
should be appended.
Hence, to get correct result, you should change the array to : {6868, 6867}.
Hope this makes things clear for you.
Cheers
Nikhil Jindal
http://sites.google.com/site/aboutnikhiljindal/
On Mon, May 30, 2011 at 3:28 PM, Bhavesh agrawal
@Naveen:
Here's a counter case:
162, 16
The next digit(2) is not greater than the last equal digit(6), still 162
comes before 16.
Here, as is done in ashu's algo, the next digit (2) should be compared with
first digit(1) and not the last equal digit(6).
Cheers
Nikhil Jindal
http
Ok. Here's a possible O(n) solution.
Assuming last digit of a is 0.
for(n=a;n=b;n+=10)
{
Calculate the sum of digits, leaving the last digit.
Find the minimum value of last digit for it to be a heavy number.
Increment count by 10-that number.
}
So here, complexity will be: O(n/10*(d-1))
where, d
.
Nikhil Jindal
https://sites.google.com/site/aboutnikhiljindal/
On Tue, Mar 15, 2011 at 4:17 PM, bittu shashank7andr...@gmail.com wrote:
U r watching an i cricket match.Suddenly the tv goes blank. Write the
steps u ll take to find the fault
purpose of this question is not to spamming but to taste
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This shld help.
Cheers
Nikhil Jindal
https://sites.google.com/site/aboutnikhiljindal/
Please
He tells the truth on tuesday.
First day is sunday.
Nice question. Took me some time to get it right.
On Mon, Feb 28, 2011 at 6:47 PM, nidhi jain nidhi.jain311...@gmail.comwrote:
Sunday,Saturday
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Algorithm Geeks
way, it doubles its speed to 2
moves per 3 second.
Hence, eventually both the robots will clash.
Cheers
Nikhil Jindal
http://www.fundoonick.blogspot.com/
On Thu, Feb 17, 2011 at 12:23 AM, bittu shashank7andr...@gmail.com wrote:
Two robots are placed at different points on a straight line
First Question:
Nt sure but shldnt t1 be greater than t2?
Second:
Since, Q is a subset of P.
P intersection Q would be Q itself.
Would be great if you can share some more questions
On Mon, Feb 14, 2011 at 7:52 PM, sankalp srivastava
richi.sankalp1...@gmail.com wrote:
First question
mode
The puzzle has recently been discussed in another thread.
On Fri, Feb 4, 2011 at 2:20 PM, bittu shashank7andr...@gmail.com wrote:
A blind man is handed a deck of 52 cards and told that exactly 10 of
these cards are facing up. How can he divide the cards into two piles,
not necessarily of
*dp(i-3)) { buff = dp(i-3);}*
*}*
@saikat: for n=10, this gives dp(10) = 20 :D
An O(n) soln.
Cheers
Nikhil Jindal
Delhi College of Engineering (DCE),
Delhi.
On Wed, Jan 19, 2011 at 10:05 PM, nishaanth nishaant...@gmail.com wrote:
How about the following dynamic programming solution.
Let dp[i
(a/b)mod m = (amodm)*(Bmodm)
where B is the multiplicative inverse of b i.e
(b*B)modM = 1 or B = 1/b
Look into the wiki page of Multiplicative inverse for more.
Hope this helps
Cheers
Nikhil Jindal
On Wed, Jan 12, 2011 at 7:06 AM, mittal mohitm.1...@gmail.com wrote:
Somehelp with (a/b)modm
On Thu, Jan 13, 2011 at 12:06 AM, Aviral Gupta aviral@gmail.com wrote:
we can do it when hcf(b,m)=1 , in that case find inverse of b by
extended euclidean mod m and then multiply it by a
Yes. And when m is prime, B(mulitplicative inverse of b) = b^(m-2)
As b^(m-1)mod m = 1 if m is prime.
Cut each slice into 8 equal pieces. Total 24 pieces.
One part consists of 3 pieces.
Total 8 parts.
On Wed, Jan 12, 2011 at 6:14 PM, bittu shashank7andr...@gmail.com wrote:
2nd puzzle
You have a birthday cake and have exactly 3 slices to cut it into 8
equal pieces. How do you do it?
Thanks
()()(), ((())), (()()), ()(()), (())()
Cheers
Nikhil Jindal
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On Tue, Sep 21, 2010 at 6:05 PM, saurabh singh saurabh.n...@gmail.comwrote:
solution 1:
use concept of quad-tree and do binary search in that tree
solution 2:
do binary search on major diagonal. ultimately u will narrow down to search
for element in 2 rows. in these two rows again do
be in the present column after(below)
the element we are on.
At max, 3n moves = O(n).
On Sun, Sep 26, 2010 at 7:34 PM, Nikhil Jindal fundoon...@yahoo.co.inwrote:
On Tue, Sep 21, 2010 at 6:05 PM, saurabh singh saurabh.n...@gmail.comwrote:
solution 1:
use concept of quad-tree and do binary search
, int item)
{
if (i0 || j0 || in-1 || jn-1)
{
printf(Not Found\n);
return;
}
if( a[i][j] == item) printf(Found: %d %d\n, i, j);
elseif( a[i][j] item) 2dsearch(i, j-1, item);
else 2dsearch(i+1, j, item);
}
Call this function as 2dsearch(0, n-1, item);
Cheers
Nikhil Jindal
Delhi
The answer would be:
(log1+1) + (log2+1) + (log3+1) + (log4+1) + ... + (log(n-1)+1)
which is equal to:
(log1+log2+log3+...+log(n-1)) + (n-1)
== *log((n-1)!) + (n-1)*
where, log everywhere is assumed to be in base 2
*This according to me will be the final answer!*
*
*
*Cheers*
*Nikhil Jindal
@vikas: Total number of elements are not n*k. Total number of elements are
n, which are divided into k lists.
@Rahul Singal: +1 for ur answer.
On Fri, Sep 17, 2010 at 12:58 PM, vikas kumar vikas.kumar...@gmail.comwrote:
@Bittu
I am confused about one point you need to process atleast n*k
Keep an augmented balanced BST. Augmented data: number of elements in the
right and left subtrees .
Insertion: logn
Find Median: logn
Cheers
Nikhil Jindal
Delhi College of Engineering
On Fri, Sep 17, 2010 at 12:09 PM, vikas kumar vikas.kumar...@gmail.comwrote:
struct list
{
median
Very Nice soln Dhritiman.
The solution to the standard LCS problem only needs O(n^2) time and O(n)
space.
And you have very intelligently solved its variation also in the same time
and space complexity.
I fell this is the correct solution.
Nikhil Jindal
On Tue, Aug 31, 2010 at 2:13 AM
@Wan:
Storing the elements as link list rather than array requires additional
space.
If we are taking extra O(n) space, then the usual approach of merging two
sorted arrays in O(n) time and memory will suffice.
On Fri, Aug 27, 2010 at 5:18 PM, Yan Wang wangyanadam1...@gmail.com wrote:
Also,
Hi Sourav,
I will first inplace sort the last √n elements in O(n) and then merge the
two sorted arrays in O(n).
The only problem: O(n) merging will not be inplace.
On Thu, Aug 26, 2010 at 5:25 PM, sourav souravs...@gmail.com wrote:
Let A[1..n] be an array such that the first (n − √n) elements
Here's my try:
The following function returns the rectangle number given the dimensions of
the rectangle.
int FindRectangleNumber(int row, int colm)
{
if (row == colm)
return row;
if (row == 1)
return colm;
if (row%2 == 0 colm%2 == 0)
return 2*FindRectangleNumber(row/2, colm/2);
if
://mc/compose?to=cho...@gmail.com
wrote:
I definitely meant a suffix Tree and not a trie. Apologize for that. :)
On Fri, Aug 20, 2010 at 8:47 PM, Nikhil Jindal
fundoon...@yahoo.co.inhttp://mc/compose?to=fundoon...@yahoo.co.in
wrote:
@chonku
As i understand, a trie is used when we have
;
}
}
tempMax-=1;
}
return 0;
}
On Sat, Aug 21, 2010 at 12:12 PM, Chonku cho...@gmail.com wrote:
I definitely meant a suffix Tree and not a trie. Apologize for that. :)
On Fri, Aug 20, 2010 at 8:47 PM, Nikhil Jindal fundoon
longest common substring in the given string.
If i get it right. You need two strings to find a common subsequence.
We use DP for it.
2010/8/18 ♪ ѕяiηivαѕαη ♪ 21.sr...@gmail.com
Example:
If my sequence is ABC..the longest common subsequence is
AC,BC,AB.
It is a very common problem...
On
dave_and_da...@juno.com wrote:
@Nikhil Jindal: What you say is true for 2 numbers, but not for more
than 2.
1 + 6 + 6 = 2 + 2 + 9 = 13, and 1 * 6 * 6 = 2 * 2 * 9 = 36.
Dave
On Aug 19, 6:00 am, Nikhil Jindal fundoon...@yahoo.co.in wrote:
Nikhil's algo is correct
:
Can we use a trie here.
Make first pass from left to right and construct the trie.
Make second pass from right to left and look for the trie branch with
maximum nodes that match the characters.
On Thu, Aug 19, 2010 at 7:46 PM, Nikhil Jindal fundoon...@yahoo.co.inwrote:
Hi All,
Givan a string
Hi All,
Givan a string, you have to find the longest palindromic substring.
For ex: Longest Palindromic substring for abclevelabc is level.
What is the most optimised solution possible?
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disclaimer:
Nikhil's algo is correct if the following is always true:
Given: x + y = S, x * y = M
and x' + y' = S', x' * y' = M'
If: S' = S and M' = M, then x = x' and y = y'
i.e for given sum and product, the elements are unique.
On Thu, Aug 19, 2010 at 12:54 PM, Nikhil Agarwal
The critical thing here is how to define your comparator function to be used
for sorting.
Sorting using the normal comparator function will return the following:
31 33 55 312
Using insertion sort O(n^2) such that the resultant concatenation is
smallest, should do it.
The steps for [55,31,312,33]
@ram:
This doesnt work for:
arr[] = {1,0,0,0}
Here, then number of 1's and 0's are not same. So the array should be left
untouched.
On Thu, Aug 19, 2010 at 7:02 PM, ram das ramnaraya...@gmail.com wrote:
{
int odd=1,even=0;
while(odd = size even =size)
{
if (
Use Knuth Shuffle algo.
On Sun, Aug 15, 2010 at 8:34 PM, Rahul Singhal nitk.ra...@gmail.com wrote:
Let X1, X2…. XN (In this case N=52) be the set of N numbers to be shuffled.
1. Set j to N
2. Generate a random number R. (uniformly distributed between 0 and 1)
3. Set k to (jR+1). k
Use hash maps...
On Mon, Aug 16, 2010 at 10:06 PM, ashita dadlani ash@gmail.com wrote:
You have a string say foobarfoo in which fo and oo aree repeated
twice.You have to find all such repeated pairs in O(n) time,The string can
only have alphanumeric elements in it.
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.
Hope this helps
Cheers
Nikhil Jindal
Delhi College of Engineering
On Thu, Aug 5, 2010 at 9:56 PM, Ashish Goel ashg...@gmail.com wrote:
can u explain how is this number reached at?
(2n)!/((n+1)!(n!))
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
) {
swap(a[k],a[n+k/2]);
} else {
swap(a[k],a[n+k/2-1]);
}
}
*Done!*
Nikhil Jindal
Delhi College of Engineering
On Sat, Aug 7, 2010 at 10:17 AM, Dave dave_and_da...@juno.com wrote:
Here's a solution that I am pretty sure is less than O(n^2). The data
are moved only once, but timing
Here's a possible O(n) soln:
for all i, I calculate a[i].diff as number of zeroes - number of ones ones
from a[0] to a[i].
I do this in O(n).
Also, I make a list of all the indexes that have a difference d(for all
possible d, which is n).
Now, for it to be possible that the number of ones and
can be missing can be found if a row in
this 2-d array remains unmodified
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Sat, Jul 31, 2010 at 10:22 PM, Nikhil Jindal
fundoon...@yahoo.co.in wrote:
At the moment, I can only think of an O
At the moment, I can only think of an O(n^3) algo.
Maybe if you can find a hash function which computes the hash value
depending on the unique characters the string contains, you can reduce it to
O(n^2).
On Sat, Jul 31, 2010 at 7:09 PM, srikanth sg srikanthini...@gmail.comwrote:
given two
For merging n companies, F(n) = n*F(n-1) for n 3.
Base case, F(3) = 3.
On Sat, Jul 31, 2010 at 6:59 PM, sourav souravs...@gmail.com wrote:
Suppose we start with n companies that eventually merge into one big
company. How many different ways are there for them to merge?
With three companies
Multiplying two n digit numbers, where multiplication of two 1 digit numbers
is O(1), is : O(n^2).
On Sat, Jul 31, 2010 at 9:12 PM, Dave dave_and_da...@juno.com wrote:
If by repeated addition method, you mean
m + m + m + ... + m (where m occurs k times)
for forming the product k*m, then the
@Ram Kumar: Yes. Simple and affective.
Just at each node:
node-left-side=node-right
node-right-side=node-side-left
i.e at each node you are setting the side of each of your child. Go on and
just do it for all nodes. Done.
On Sat, Jul 31, 2010 at 9:39 AM, Ram Kumar
guess the same number every time.
For the given puzzle, all men guess the same number and at least one of them
will be correct. :)
Nikhil Jindal
Department of Computer Engineering
Delhi College of Engineering http://www.dce.edu, Delhi
My Blog: http://fundoonick.blogspot.com
My LinkedIn Profile: http
of length 4).
PS: I am assuming by maximum subsequence, you meant longest.
Nikhil Jindal
On Sun, Jul 4, 2010 at 3:21 PM, jalaj jaiswal jalaj.jaiswa...@gmail.comwrote:
You are given an array ' containing 0s and 1s. Find O(n) time and O(1)
space algorithm to find the maximum sub sequence which has equal
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